Lời giải chi tiết:

* Giả sử \(A\left( {a;0;0} \right),\,\,B\left( {0;b;0} \right),\,\,C\left( {0;0;c} \right)\,\,\left( {DK:\,\,a,b,c > 0} \right)\).

\( \Rightarrow Pt\left( P \right):\,\,\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\)

* \(M\left( {4;1;2} \right) \in \left( P \right) \Rightarrow \dfrac{4}{a} + \dfrac{1}{b} + \dfrac{2}{c} = 1\,\,\left( 1 \right)\)

* Áp dụng BĐT Cô-si ta có: \(\dfrac{4}{a} + \dfrac{1}{b} + \dfrac{2}{c} \ge 3\sqrt[3]{{\dfrac{4}{a}.\dfrac{1}{b}.\dfrac{2}{c}}} \Rightarrow 1 \ge \dfrac{6}{{\sqrt[3]{{abc}}}} \Rightarrow abc \ge {6^3}\)

* \({V_{OABC}} = \dfrac{1}{6}abc \Rightarrow {V_{OABC\,\,\min }} = \dfrac{1}{6}{.6^3} = 36\).

Dấu "=" xảy ra \( \Leftrightarrow \dfrac{4}{a} = \dfrac{1}{b} = \dfrac{2}{c} = \dfrac{1}{3} \Rightarrow \left\{ \begin{array}{l}a = 12\\b = 3\\c = 6\end{array} \right. \Rightarrow \left( P \right):\,\,\dfrac{x}{{12}} + \dfrac{y}{3} + \dfrac{z}{6} = 1\).

Chọn A.